NCERT Solutions for Class 11 Chemistry Chapter 7 Equilibrium

Ans : (a) If the volume of the container is suddenly increased, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume. 
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. 
Hence, the rate of condensation decreases initially. 
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system. 

Ans : The equilibrium constant (K,) for the give reaction is: 
\(\begin{aligned} \mathrm{K}_{c} &=\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]} \\ &=\frac{(1.90)^{2} \mathrm{M}^{2}}{(0.60)^{2}(0.821) \mathrm{M}^{3}} \\ &=12.239 \mathrm{M}^{-1}(\text { approximately }) \end{aligned}\)
Hence, \(K_{c} \text { for the equilibrium is }^{12.239 \mathrm{M}^{-1}}\) 

Ans : \(\begin{array}{l}{\text { Partial pressure of I atoms, }} \\ {p_{1}=\frac{40}{100} \times p_{\text { lotalal }}} \\ {=\frac{40}{100} \times 10^{5}} \\ {=4 \times 10^{4} \mathrm{Pa}} \\ {\text { Partial pressure of I_ } \mathrm{I}_{2} \text { molecules, }}\end{array}\)
\(\begin{array}{l}{p_{1_{2}}=\frac{60}{100} \times p_{\text { total }}} \\ {=\frac{60}{100} \times 10^{5}} \\ {=6 \times 10^{4} \mathrm{Pa}} \\ {\text { Now, for the given reaction, }}\end{array}\)
\(\begin{aligned} K_{p} &=\frac{(p \mathrm{I})^{2}}{p_{1_{2}}} \\ &=\frac{\left(4 \times 10^{4}\right)^{2} \mathrm{Pa}^{2}}{6 \times 10^{4} \mathrm{Pa}} \\ &=2.67 \times 10^{4} \mathrm{Pa} \end{aligned}\) 

Ans : (i) \(\quad K_{e}=\frac{\left[\mathrm{NO}_{(g)}\right]^{2}\left[\mathrm{Cl}_{2(\mathrm{g}\}}\right]}{\left[\mathrm{NOCl}_{(g)}\right]^{2}}\)
(ii) \(K_{c}=\frac{\left[\mathrm{CuO}_{(s)}\right]^{2}\left[\mathrm{NO}_{2(\mathrm{g})}\right]^{4}\left[\mathrm{O}_{2(\mathrm{g})}\right]}{\left[\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2(\mathrm{i})}\right]^{2}}\)
\(=\left[\mathrm{NO}_{2(\mathrm{g})}\right]^{4}\left[\mathrm{O}_{2(\mathrm{g})}\right]\)
 

Ans : \(\begin{array}{l}{\text { The relation between } \mathrm{K}_{\mathrm{p}} \text { and } \mathrm{K}_{\mathrm{c}} \text { is given as: }} \\ {\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}}} \\ {\text { (a) Here, }} \\ {\Delta \mathrm{n}=3-2=1} \\ {\mathrm{R}=0.0831 \text { barLmol-1 } \mathrm{K}^{-1}} \\ {\mathrm{T}=500 \mathrm{K}} \\ {\mathrm{K}_{\mathrm{p}}=1.8 \times 10^{-2}}\end{array}\) 
\(\begin{array}{l}{\mathrm{K} \mathrm{ow},} \\ {\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}}} \\ {\Rightarrow 1.8 \times 10^{-2}=K_{c}(0.0831 \times 500)^{1}} \\ {\Rightarrow K_{c}=\frac{1.8 \times 10^{-2}}{0.0831 \times 500}} \\ {=4.33 \times 10^{-4}(\text { approximately })}\end{array}\)
\(\begin{array}{l}{\text { (b) Here, }} \\ {\Delta n=2-1=1} \\ {R=0.0831 \text { barLmol }^{-1} K^{-1}} \\ {T=1073 \mathrm{K}} \\ {K_{p}=167} \\ {\text { Now, }} \\ {K_{p}=K_{c}(R T)^{\Delta n}}\end{array}\)
\(\begin{aligned} \Rightarrow 167 &=K_{c}(0.0831 \times 1073)^{\Delta n} \\ \Rightarrow K_{c} &=\frac{167}{0.0831 \times 1073} \\ &=1.87(\text { approximately }) \end{aligned}\)